Kc Calculation for BrF5 ↔ Br2 + 5F2 Reaction

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To calculate the equilibrium constant Kc for the reaction 2BrF5 ↔ Br2 + 5F2, the correct setup is to place the products on the right side of the equation. This means Kc is expressed as [Br2][F2]^5 / [BrF5]^2. By convention, the right-hand side (RHS) represents the products, while the left-hand side (LHS) represents the reactants. The forward reaction is defined as LHS to RHS, and the reverse reaction is RHS to LHS, with the latter corresponding to the rate-limiting step in this context.
Mag
2BrF5 ↔ Br2 + 5F2
If we wanted to solve for Kc (Kc=product/reactants) would we assume that the products are on the right side of the equation. In other words would we set it up like this

[BrF5]^2 / [Br2] [F2]^5 or


[Br2][F2]^5 / [BrF5]^2
 
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The latter. Yes, by convention, the RHS contains the products.

By this convention, LHS -> RHS is considered the forward direction/reaction and RHS -> LHS is considered the reverse reaction.
 
assuming that the latter corresponds to the rate limiting reaction...yes.


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